Question

 The area of the region given by A={(x,y); \(x^2\)≤y≤min{x+2,4−3x}} is

• A. \(\frac{31}{8}\)
• B. \(\frac{17}{6}\)
• C. \(\frac{19}{6}\)
• D. \(\frac{27}{8}\)

Answer 1

The Correct Option is B

A = {(x, y) : x² ≤ y ≤  min {x + 2, 4 – 3x}
So, the area of the required region
A=\(\int_{-1}^{\frac{1}{2}}\)(x+2−x²)dx+\(\int_{1}^{\frac{1}{2}}\)(4−3x−x²)dx
=[\(\frac{x^2}{2}+2x-\frac{x^3}{3}\)]\(^{\frac{1}{2}}\)_-1+[4x−\(\frac{3x^2}{2}\)−\(\frac{x^3}{3}\)]\(^{\frac{1}{2}}\)₁
=(\(\frac{1}{8}\)+1−\(\frac{1}{24}\))−(\(\frac{1}{2}\)−2+\(\frac{1}{3}\))+(4−\(\frac{3}{2}\)−\(\frac{1}{3}\))−(2−\(\frac{3}{8}\)−\(\frac{1}{24}\))=\(\frac{17}{6}\)